Problem 5.65 - Problem 5.65 Given: Find: [Difficulty: 3]...

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Problem 5.65 [Difficulty: 3] Given: Air flow through porous surface into narrow gap Find: The acceleration vector would be: Solution: We will apply the continuity equation to the control volume shown: CS CV A d V V d t 0 Governing Equations: (Continuity) t V z V w y V v x V u Dt V D a p (Particle Accleration) Assumptions: (1) Steady flow (2) Incompressible flow (3) Uniform flow at every section Based on the above assumptions the continuity equation reduces to: 0x w v 0 hw ux () Solving for u: () v 0 x h We apply the differential form of continuity to find v: x u y v 0 x u v 0 h y v Therefore the y-velocity v is: vy v 0 h d fx v 0 y h Now at y = 0: vv 0 Therefore we can solve for f(x): v 0 v 0 0 h 0 So we find that the y-component of velocity is: 0 1 y h y V v x V u a p Based on the above assumptions the particle acceleration reduces to:
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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